Homework Help: Vector geometry: Proof of a trapezium and cross and dot product help

1. Nov 3, 2012

FilipaE

1. The problem statement, all variables and given/known data
#1 Given that the angle between the vectors a and b is 2Pi/3 and |a|=3 and |b|=4 calculate:
(axb)^2 [(2a+b)x(a+2b)]^2

#2 Given three unit vectors, a, b, c such that a+b+c=0 find (a dot b) + (b dot c) + (c dot a)

#3 Given AB=a+2b BC=-4a-b CD= -5a-3b where a and b are any two vectors, prove that ABCD is a trapezium

2. Relevant equations

Given above

3. The attempt at a solution
#1 I know that axb = |a||b|sinx
and axa = 0 and axb = -b x a but i do not understand how the powers or constants affect this?

#2 No idea

#3 A trapezium has one set of parallel sides only and i know if vectors are parallel they are a multiple of each other, however none of these seem to be multiples of each other?

2. Nov 3, 2012

micromass

This is false. The length of $a\times b$ equals $|a||b|\sin\theta$.

Anyway, what do you mean with $(a\times b)^2$?? What does it mean to take the square of a vector??

3. Nov 3, 2012

FilipaE

I am using x as the cross product, meaning the cross product of a and b is |a||b|sinx, is this not correct?

4. Nov 3, 2012

FilipaE

and (a x b)^2 means the cross product of a and b all sqaured

5. Nov 3, 2012

micromass

No, that is not correct. Review your definition of cross product.

6. Nov 3, 2012

micromass

Yes, I get that: it means that you take the square of the cross product. But the cross product is a vector. What do you mean with the square of a vector?

7. Nov 3, 2012

FilipaE

This is the definition of cross product online |a x b|=|a||b|sinx
I have no idea what you mean vectors is a very confusing topic for me

8. Nov 3, 2012

micromass

Yes, the definition $|a\times b|=|a||b|\sin x$ is correct. But writing $a\times b=|a||b|sin x$ is not correct. Do you see the difference?

9. Nov 3, 2012

FilipaE

Oh i see, so how does the formula differ without the modulus sign? and what happens where there is addition involved as in Q1 part 2

10. Nov 3, 2012

micromass

Do you understand that $a\times b$ is a vector and that $|a\times b|$ is a scalar??

11. Nov 3, 2012

FilipaE

yes i do now, but given this information how would i write the vector a x b?

12. Nov 3, 2012

micromass

Before you do that, you got to understand what $(a\times b)^2$ is. What does it mean to square a vector?

13. Nov 3, 2012

FilipaE

Is it the dot product of the two vectors, creating a scalar?

14. Nov 3, 2012

micromass

Yes.

So you need to find $(a\times b)\cdot (a\times b)$.

Given a vector v, what is the definition of v.v?

15. Nov 3, 2012

FilipaE

|v|^2?

16. Nov 3, 2012

micromass

So $(a\times b)\cdot (a\times b)=|a\times b|^2$. Now you should be able to calculate this.

17. Nov 3, 2012

FilipaE

Thankyou! I have got the answer 108?

So for [(2a+b) x(a+2b)]^2 I do [(2a+b)x(a+2b)] dot [(2a+b)x(a+2b)] = |(2a+b)x(a+2b)|^2

I know to do this i need to use the distributive law. Is it

2a x a + 2a x2b + b xa + b x2b?
Where 2a x a = 0 and bx2b=0?

18. Nov 3, 2012

FilipaE

So the answer is 108? Thankyou!
For [(2a+b) x(a+2b)]^2 is this the same as
(2a+b) x(a+2b) dot (2a+b) x(a+2b) = |(2a+b) x(a+2b)|^2 ??

I know i have to use some sort of distributive law here but i am not sure on how to do it would it be 2a x a + 2a x 2b + b x a + b x 2b ? where 2a x a = 0 and b x 2b =0??

19. Nov 3, 2012

FilipaE

Sorry didnt think the first message had sent!